Summary

You searched for: inst=10/7

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1

New Number: 5.5 |  AESZ: 22  |  Superseeker: 10/7 295/7  |  Hash: 5b96eae0872756be1130d4b12ffe60a6  

Degree: 5

\(7^{2} \theta^4-7 x\left(155\theta^4+286\theta^3+234\theta^2+91\theta+14\right)-x^{2}\left(16105\theta^4+68044\theta^3+102261\theta^2+66094\theta+15736\right)+2^{3} x^{3}\left(2625\theta^4+8589\theta^3+9071\theta^2+3759\theta+476\right)-2^{4} x^{4}\left(465\theta^4+1266\theta^3+1439\theta^2+806\theta+184\right)+2^{9} x^{5}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 2, 34, 488, 9826, ...
--> OEIS
Normalized instanton numbers (n0=1): 10/7, 65/7, 295/7, 3065/7, 4245, ... ; Common denominator:...

Discriminant

\((32z-1)(z^2-11z-1)(4z-7)^2\)

Local exponents

\(\frac{ 11}{ 2}-\frac{ 5}{ 2}\sqrt{ 5}\)\(0\)\(\frac{ 1}{ 32}\)\(\frac{ 7}{ 4}\)\(\frac{ 11}{ 2}+\frac{ 5}{ 2}\sqrt{ 5}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(1\)\(0\)\(1\)\(1\)\(1\)\(1\)
\(1\)\(0\)\(1\)\(3\)\(1\)\(1\)
\(2\)\(0\)\(2\)\(4\)\(2\)\(1\)

Note:

There is a second MUM-point at infinity, corresponding to Operator AESZ 118/5.16
A-Incarnation: five (1,1) sections in ${\bf P}^4 \times {\bf P}^4$.Quotient by ${\bf Z}/2$ of this:
the Reye congruence Calabi-Yau threefold.

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2

New Number: 6.36 |  AESZ:  |  Superseeker: 10/7 508/7  |  Hash: b890cacbc73012eb6554263c3ea04707  

Degree: 6

\(7^{2} \theta^4-2 7 x\left(60\theta^4+24\theta^3-9\theta^2-21\theta-7\right)-2^{2} x^{2}\left(6492\theta^4+30192\theta^3+46665\theta^2+30786\theta+7777\right)+2^{4} x^{3}\left(3632\theta^4-27552\theta^3-133920\theta^2-173880\theta-76083\right)+2^{9} x^{4}\left(1776\theta^4+10272\theta^3+15264\theta^2+7608\theta+121\right)-2^{14} x^{5}\left(48\theta^4-480\theta^3-2016\theta^2-2568\theta-1091\right)-2^{19} x^{6}\left((2\theta+3)^4\right)\)

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Coefficients of the holomorphic solution: 1, -2, 38, 204, 7462, ...
--> OEIS
Normalized instanton numbers (n0=1): 10/7, 100/7, 508/7, 808, 59910/7, ... ; Common denominator:...

Discriminant

\(-(16z+1)(32z-1)(4z+1)^2(32z-7)^2\)

Local exponents

\(-\frac{ 1}{ 4}\)\(-\frac{ 1}{ 16}\)\(0\)\(\frac{ 1}{ 32}\)\(\frac{ 7}{ 32}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 3}{ 2}\)
\(-\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(\frac{ 3}{ 2}\)
\(1\)\(1\)\(0\)\(1\)\(3\)\(\frac{ 3}{ 2}\)
\(\frac{ 3}{ 2}\)\(2\)\(0\)\(2\)\(4\)\(\frac{ 3}{ 2}\)

Note:

This is operator "6.36" from ...

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3

New Number: 7.1 |  AESZ:  |  Superseeker: 10/7 508/7  |  Hash: 08ab3cb496250adfa30bc3e24ac63c4f  

Degree: 7

\(7^{2} \theta^4-2 7 x\theta(46\theta^3+52\theta^2+33\theta+7)-2^{2} x^{2}\left(7332\theta^4+28848\theta^3+42633\theta^2+26670\theta+6272\right)-2^{4} x^{3}\left(2860\theta^4+44760\theta^3+120483\theta^2+111279\theta+35098\right)+2^{9} x^{4}\left(2230\theta^4+5920\theta^3-741\theta^2-6509\theta-3049\right)+2^{14} x^{5}\left(174\theta^4+1320\theta^3+1971\theta^2+1095\theta+190\right)-2^{19} x^{6}\left(22\theta^4+24\theta^3-9\theta^2-21\theta-7\right)-2^{25} x^{7}\left((\theta+1)^4\right)\)

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Coefficients of the holomorphic solution: 1, 0, 32, 288, 7776, ...
--> OEIS
Normalized instanton numbers (n0=1): 10/7, 100/7, 508/7, 808, 59910/7, ... ; Common denominator:...

Discriminant

\(-(16z+1)(32z-1)(32z-7)^2(4z+1)^3\)

Local exponents

\(-\frac{ 1}{ 4}\)\(-\frac{ 1}{ 16}\)\(0\)\(\frac{ 1}{ 32}\)\(\frac{ 7}{ 32}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(1\)
\(\frac{ 1}{ 2}\)\(1\)\(0\)\(1\)\(1\)\(1\)
\(\frac{ 3}{ 2}\)\(1\)\(0\)\(1\)\(3\)\(1\)
\(2\)\(2\)\(0\)\(2\)\(4\)\(1\)

Note:

This is operator "7.1" from ...

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